OH- Lewis structure

OH^– (hydroxide) has one oxygen atom and one hydrogen atom.

In the OH^– Lewis structure, there is a single bond between the oxygen and hydrogen atom, and on the oxygen atom, there are three lone pairs.

Also, there is a negative (-1) charge on the oxygen atom.

Alternative method: Lewis structure of OH^–

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, oxygen lies in group 16, and hydrogen lies in group 1.

Hence, oxygen has six valence electrons and hydrogen has one valence electron.

Since OH^– has one oxygen atom and one hydrogen atom, so…

Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of one hydrogen atom = 1 × 1 = 1

Now the OH^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 6 + 1 + 1 = 8

Learn how to find: Oxygen valence electrons and Hydrogen valence electrons

• Second, find the total electron pairs

We have a total of 8 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 8 ÷ 2 = 4

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Hence, here we have to assume that the central atom is oxygen.

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 4 electron pairs. And one O — H bond is already marked. So we have to only mark the remaining three electron pairs as lone pairs on the sketch.

Also remember that oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atom is hydrogen. But no need to mark on hydrogen, because hydrogen already has two electrons.

So for oxygen, there are three lone pairs.

Mark the lone pair on the sketch as follows:

Formal charge

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

For hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

Here, both oxygen and hydrogen atoms have charges, so mark them on the sketch as follows:

Final structure

The final structure of OH^– comprises a single oxygen atom connected to one hydrogen atom through a covalent bond. In this configuration, the oxygen atom satisfies the octet rule by maintaining three lone pairs alongside its single shared bond, while the hydrogen atom achieves its stable duet state. This arrangement represents the most stable state for the ion because it results in a formal charge of -1 on the oxygen atom and a formal charge of zero on the hydrogen atom. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of the hydroxide ion.

To properly represent this as a polyatomic ion, the entire Lewis structure is enclosed within square brackets. The overall charge of 1- is then written as a superscript outside the brackets at the top right, indicating that the structure possesses one additional electron beyond the valence count of the neutral atoms.

Next: N₂H₂ Lewis structure

External video

• OH- Lewis Structure – How to Draw the Lewis Dot Structure for the Hydroxide Ion – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/OHMinus-Lewis-Structure.html
• https://www.chemistryscl.com/general/OH–lewis-structure/index.php
• https://lambdageeks.com/oh-lewis-structure/
• https://socratic.org/questions/how-do-you-write-the-electron-dot-diagram-for-the-hydroxide-ion-oh